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The Skunk Works.

Lockheed-Martin's Advanced Development Programs.

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What is the speed of a B-2 plane?

The B-2 Spirit, a stealth bomber used by the United States Air Force, has a maximum speed of approximately 630 miles per hour (1,015 kilometers per hour) at high altitude. Its operational range is about 6,000 miles (9,656 kilometers) without refueling, allowing it to carry out long missions. The aircraft's design emphasizes stealth capabilities, enabling it to evade radar detection while delivering precision strikes.


How do the squares of sums and differences relate?

(a + b) = a^2 + 2ab + b^2 (a - b)^2 = a^2 - 2ab + b^2 or you can work like this: [a + (-b)]^2 = a^2 + 2a(-b) + (-b)^2 (a - b)^2 = a^2 - 2ab + b^2


What are the different special product formulas?

1. Square of a binomial (a+b)^2 = a^2 + 2ab + b^2 carry the signs as you solve 2. Square of a Trinomial (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc carry the sings as you solve 3. Cube of a Binomial (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3 4. Product of sum and difference (a+b)(a-b) = a^2 - b^2 5. Product of a binomial and a special multinomial (a+b)(a^2 - ab + b^2) = a^3-b^3 (a-b)(a^2 + ab + b^2) = a^3-b^3


If a squared is less than b squared is a less than b?

If ( a^2 < b^2 ), it does not necessarily mean that ( a < b ). This is because both ( a ) and ( b ) could be negative numbers. For example, if ( a = -3 ) and ( b = 2 ), then ( a^2 = 9 ) and ( b^2 = 4 ), so ( a^2 > b^2 ). Therefore, while ( a^2 < b^2 ) could imply ( |a| < |b| ), it does not definitively indicate that ( a < b ).


What is proof of Heron's Formula?

This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)